Re: tensor product twister


Jannick Asmus wrote:
On 09.11.2006 05:08, oferlock@xxxxxxxxx wrote:
If A and B are algebras (over an algebraically closed field F) with no
nilpotent elements, can we say that A*B (tensor product) has no
nilpotent elements? I thought that should be easy, but can not say that
since linear combination of tensors can still be nilpotent. Is there a
theorem that rules that out?

Reduce to the case where A and B are finitely generated F-algebras. Then
use the assumption on F and the Hilbert Nullstellensatz that every
maximal ideal M of A*B has the unique representation m1*B+A*m2 where m1
and m2 are maximal ideals in A and B, respectively. Now try to deduce
the claim from the fact that in every finitely generated algebra over a
field the nilradical is the intersection of all maximal ideals (BTW:
This is some version of the Hilbert Nullstellensatz, too).

HTH.

J.

it's very interesting approach, and works, but i think you might have
to replace maximal with prime.

.

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